In △ ABC, the opposite side lengths of internal angles a, B and C are a, B and C respectively. If a2-c2 = 2B and sinacosc = 3cosasinc, then B = () A. 4 B. 4 two C. 2 three D. 3 three

In △ ABC, the opposite side lengths of internal angles a, B and C are a, B and C respectively. If a2-c2 = 2B and sinacosc = 3cosasinc, then B = () A. 4 B. 4 two C. 2 three D. 3 three

sinAcosC=3cosAsinC，
The sine and cosine theorems are used to obtain:
aa2+b2−c2
2ab＝3cb2+c2−a2
2bc
The solution is: 2 (a2-c2) = B2 ①
Because: a2-c2 = 2B ②
From ① and ②: B = 4
Therefore: a

In △ ABC, the opposite side lengths of internal angles a, B and C are a, B and C respectively. If a2-c2 = 2B and sinacosc = 3cosasinc, then B = () A. 4 B. 4 two C. 2 three D. 3 three

sinAcosC=3cosAsinC，
The sine and cosine theorems are used to obtain:
aa2+b2−c2
2ab＝3cb2+c2−a2
2bc
The solution is: 2 (a2-c2) = B2 ①
Because: a2-c2 = 2B ②
From ① and ②: B = 4
Therefore: a

In the known triangle ABC, ab = 10, BC = 9, AC = 17, find the height ad on the BC edge

From the cosine theorem, the cos angle BAC = (21 * 21-17 * 17-10 * 10) / 2 * 10 * 17 = - 52 / 340 is obtuse, so D is on BC, excluding the possibility of BC extension line. Let BD = x, DC = y, x + y = 21 and ad = S. according to the Pythagorean theorem, the square of 17 * 17-x = 10 * 10-y, the square of X - y = 1

In triangle ABC, the length of C is twice the root sign two, Tana = 3, tanb = 2. Try to find the area of a, B and triangle ABC

tan（A+B）=-tanC
（tanA+tanB）/（1-tanA*tanB）=-tanC
tanC=1
C=45
Tanb = SINB / CoSb = SINB / radical [1 - (SINB) ^ 2]
tanB^2=sinB^2/[1-(sinB)^2]=4
SINB = 2 radical 5 / 5
Similarly, Sina = 3 root sign 10 / 10
c/sinC=b/sinB=a/sinA
A = 6, root number 10 / 5
B = 8 radical 5 / 5
S=(1/2)absinC=24/5

In the triangle ABC, a / C = (root 3) - 1, tanb / Tanc = (2a-c) / C, find the angles a, B, C?

Tanb / Tanc = (2a-c) / C = (2sina sinc) / sinc, that is, SINB * COSC = 2sina * CoSb sinc * CoSb, so the shift term uses the sum angle formula of sine to get sin (B + C) = 2sina * CoSb = Sina, so CoSb = 1 / 2, so B = 60, and Sina / sinc = root 3-1, so sin (120-c) / sinc = root 3-1, so cot

In triangle ABC, A.B.C satisfies the square of B + the square of C BC = the square of A. C / b = 1 / 2 + root 3, find ∠ A. find tanb

1. B ^ 2 + C ^ 2-bc = a ^ 2 shift B ^ 2 + C ^ 2-A ^ 2 = BC (b ^ 2 + C ^ 2-A ^ 2) / 2BC = 1 / 2 (b ^ 2 + C ^ 2-A ^ 2) / 2BC = cos60, so ∠ a = 60C / B = 1 / 2 + √ 3C / b = (1 + 2 √ 3) / 2 set C = (1 + 2 √ 3) XB = 2x (b ^ 2 + C ^ 2-A ^ 2) / 2BC = 1 / 2 solution a = √ 150.5 √ 3 / √ 15 = SINB / 2sinb = √ 5 / 5cosb = 2 √ 5 / 5